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10.4 - Ellipses

10.4 - Ellipses. Ellipses - Warm Up. Solve each equation. 1. 27 = x 2 + 11 2. x 2 = 48 3. 84 = 120 – x 2. 1. 27 = x 2 + 11 16 = x 2 x = ± 16 = ±4 2. x 2 = 48 x = ± 48 x = ± 4 3 3. 84 = 120 – x 2 –36 = – x 2

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10.4 - Ellipses

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  1. 10.4 - Ellipses

  2. Ellipses - Warm Up Solve each equation. 1. 27 = x2 + 11 2.x2 = 48 3. 84 = 120 – x2

  3. 1. 27 = x2 + 11 16 = x2 x = ± 16 = ±4 2.x2 = 48 x = ± 48 x = ± 4 3 3. 84 = 120 – x2 –36 = –x2 36 = x2 x = ± 36 = ±6 Solutions Ellipses – Warm Up

  4. ELLIPSE TERMS Vertex Co-vertex Focus (h, k ) a V=(h , k) c (h ± a , k) b (h, k ± b) horizontal Co-vertex 2a Vertex Major axis vertical 2b Minor axis (h ± c , k)

  5. ELLIPSE TERMS Vertex Minor axis a Co-vertex (h, k ) V=(h , k) (h, k ± a) Co-vertex b c (h ± b , k) Focus vertical Vertex 2a Major axis horizontal 2b (h , k ± c )

  6. CONVERTING to STANDARD FORM • x² + 4y² + 4x – 24y + 24 = 0 • Groups the x terms and y terms • x² + 4x + 4y² – 24y + 24 = 0 • Complete the square • x² + 4x + 4(y² – 6y) + 24 = 0 • x² + 4x + 4 + 4(y² – 6y + 9) = -24 + 4 +36 • (x + 2)² + 4(y – 3)² = 16 • Divide to put in standard form • (x + 2)²/16 + 4(y – 3)²/16 = 1

  7. x2 b2 y2 a2 + = 1 Standard form for an equation of an ellipse with a vertical major axis. x2 9 y2 16 + = 1 Substitute 9 for b2 and 16 for a2. x2 9 y2 16 An equation of the ellipse is + = 1. Writing an Equation of an Ellipse Write an equation in standard form of an ellipse that has a vertex at (0, –4), a co-vertex at (3, 0), and is centered at the origin. Since (0, –4) is a vertex of the ellipse, the other vertex is at (0, 4), and the major axis is vertical. Since (3, 0) is a co-vertex, the other co-vertex is at (–3, 0), and the minor axis is horizontal. So, a = 4, b = 3, a2 = 16, and b2 = 9.

  8. Graph and Label • b) Find coordinates of vertices, covertices, foci • Center = (-3,2) • Horizontal ellipse since the a² value is under x terms • Since a = 3 and b = 2 • Vertices are 3 points left and right from center  (-3 ± 3, 2) • Covertices are 2 points up and down  (-3, 2 ± 2) • Now to find focus points • Use c² = a² - b² • So c² = 9 – 4 = 5 • c² = 5 and c = ±√5 • Focus points are √5 left and right from the center  F(-3 ±√5 , 2) • a) GRAPH • Plot Center (-3,2) • a = 3 (go left and right) • b = 2 (go up and down)

  9. Graph and Label • b) Find coordinates of vertices, covertices, foci • Center = (3,-1) • Vertical ellipse since the a² value is under y terms • Since a = 4 and b = 2 • Vertices are 3 points up and down from center  (3, -1 ± 2) • Covertices are 2 points left and right  (3 ± 2, -1) • Now to find focus points • Use c² = a² - b² • So c² = 16 – 4 = 12 • c² = 12 and c = ±2√3 • Focus points are 2√3 up and down from the center  F(3,-1 ±2√3) • a) GRAPH • Plot Center (3,1) • a = 4 (go up and down) • b = 2 (go left and right)

  10. Write the equation of the ellipse given…endpoints of major axis are at (-11,5) and (7,5)endpoints of minor axis are at (-2,9) and (-2,1) • Draw a graph with given info • Use given info to get measurements • Major axis = 2a • Major axis is 18 units, • so a = 9 • Minor axis = 2b • Minor axis is 8 units, • so b = 4 • Use standard form • Need values for h,k, a and b • We know a = 9 and b = 4 • How do we find center??? • Use midpoint formula • (h , k) = (-2 , 5) • Plug into formula C = (-2,9) A = (-11,5) B = (7,5) major minor D = (-2,1)

  11. + = 1 Standard form for an ellipse with a horizontal major axis. x2 a2 y2 b2 x2 100 x2 100 y2 25 y2 25 + = 1 Substitute 100 for a2 and 25 for b2. An equation of the ellipse is + = 1. Let’s Try One Find an equation of an ellipse centered at the origin that is 20 units wide and 10 units high. Since the largest part of the ellipse is horizontal and the width is 20 units, place the vertices at (±10, 0). Place the co-vertices at (0, ±5). So, a = 10, b = 5, a2 = 100, and b2 = 25.

  12. Write the equation of the ellipse given…foci are at (2,5) and (-2,5)vertex at (-3,5) • Draw a graph with given info • Use given info to get measurements • Find the center first. • The center is in the middle of the foci. Use midpoint formula to find (h , k) • (h , k) = (0 , 5) • Then c = distance from center to foci • So c = 2 • Then a = distance from center to vertex • so a = 3 • Use standard form • Need values for h,k, a and b • We know a = 3, c = 2, (h,k) = (0,5) • How do we find b??? • Use c² = a² – b² • 4 = 9 – b² • we get b² = 5 • Plug into formula a = 3 c = 2 (h,k) = (0,5)

  13. 9x2 + y2 = 36 + = 1 Write in standard form. Since 36 > 4 and 36 is with y2, the major axis is vertical, a2 = 36, and b2 = 4. c2 = a2 – b2Find c. = 36 – 4 Substitute 4 for a2 and 36 for b2. = 32 x2 y2 4 36 c = ± 32 = ± 4 2 The major axis is vertical, so the coordinates of the foci are (0, ±c). The foci are: (0, 4 2 ) and (0, – 4 2). Working Backwards Find the foci of the ellipse with the equation 9x2 + y2 = 36. Graph the ellipse.

  14. Since c = 4 and b = 2, c2 = 16, and b2 = 4. c2 = a2 – b2Use the equation to find a2. 16 = a2 – 4 Substitute 16 for c2 and 4 for b2. a2 = 20 Simplify. x2 x2 y2 y2 4 4 + = 1 Substitute 20 for a2 and 4 for b2. 20 20 An equation of the ellipse is + = 1. Let’s Try One Write an equation of the ellipse with foci at (0, ±4) and co-vertices at (±2, 0). Since the foci have coordinates (0, ±4), the major axis is vertical.

  15. Application of foci in ellipses • You may think that most objects in space that orbit something else move in circles, but that isn't the case. Although some objects follow circular orbits, most orbits are shaped more like "stretched out" circles or ovals. Mathematicians and astronomers call this oval shape an ellipse. The Sun isn't quite at the center of a planet's elliptical orbit. An ellipse has a point a little bit away from the center called the "focus". The Sun is at the focus of the ellipse. Because the Sun is at the focus, not the center, of the ellipse, the planet moves closer to and further away from the Sun every orbit

  16. More on orbits… • Orbits are ellipses. An ellipse can be like a circle, or it can be long and skinny. Mathematicians and astronomers use the term "eccentricity" to describe the shape of an orbit. Eccentricity = c/a. An orbit shaped almost like a circle has a low eccentricity close to zero. A long, skinny orbit has a high eccentricity, close to one • http://windows2universe.org/physical_science/physics/mechanics/orbit/orbit_shape_interactive.html

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